Problem 26 - Algebra A Combined Function

Question

Solve each system. $$ \left\\{\begin{aligned} 3 x-3 y+z &=-1 \\ 3 x-y-z &=3 \\ -6 x+4 y+3 z &=-8 \end{aligned}\right. $$

Step-by-Step Solution

Answer

\(x = \frac{1}{3}, y = 0, z = -2\).

Step 1: Label the Equations

We begin by labeling the equations for reference:1. \(3x - 3y + z = -1\)2. \(3x - y - z = 3\)3. \(-6x + 4y + 3z = -8\)

Step 2: Eliminate a Variable

Let's eliminate the variable \(z\) from equations 1 and 2. Add equation 1 and equation 2:\[(3x - 3y + z) + (3x - y - z) = -1 + 3\]This simplifies to:\[6x - 4y = 2 \;\text{(Equation 4)}\]

Step 3: Eliminate the Same Variable Again

Next, eliminate \(z\) between equation 2 and equation 3. Multiply equation 2 by 3 and add it to equation 3:\[3(3x - y - z) = 3(3)\]\[9x - 3y - 3z = 9\]Now add to equation 3:\[(9x - 3y - 3z) + (-6x + 4y + 3z) = 9 + (-8)\]This simplifies to:\[3x + y = 1 \;\text{(Equation 5)}\]

Step 4: Solve the Resulting System of Two Equations

We now have two equations without \(z\):1. \(6x - 4y = 2\)2. \(3x + y = 1\)Multiply the second equation by 4:\[12x + 4y = 4\]Add this to equation 4:\[(6x - 4y) + (12x + 4y) = 2 + 4\]This simplifies to:\[18x = 6\]Solve for \(x\):\[x = \frac{1}{3}\]

Step 5: Substitute Back to Find Another Variable

Substitute \(x = \frac{1}{3}\) into equation 5:\[3\left(\frac{1}{3}\right) + y = 1\]Simplifies to:\[1 + y = 1\]Solving gives:\[y = 0\]

Step 6: Solve for the Remaining Variable

We now have \(x = \frac{1}{3}\) and \(y = 0\). Substitute these into equation 2 to find \(z\):\[3\left(\frac{1}{3}\right) - 0 - z = 3\]\[1 - z = 3\]Solving gives:\[z = -2\]

Step 7: Present the Solution

The solution to the system of equations is:\[x = \frac{1}{3},\; y = 0,\; z = -2\].

Key Concepts

Elimination Method

The elimination method is a strategic approach to solving systems of linear equations. By performing operations such as addition or subtraction on the equations, we can eliminate one or more variables, simplifying the system. This method relies on generating new equations that aren't as complex as the original ones.

• Identify the variable you want to eliminate. Often, it's best to choose one that appears with similar coefficients across the equations.
• Manipulate the equations algebraically, such as by multiplying or dividing them by constants, so the coefficients of the chosen variable are opposites. This allows them to cancel out when the equations are added or subtracted.
• Once a variable is eliminated, solve the resulting simpler system for the remaining variables.

In our exercise, we eliminated the variable \( z \) by combining the equations strategically. This method simplifies the system into a two-variable problem, making it easier to solve algebraically.

Linear Equations

Linear equations are the backbone of systems like the one given in this exercise. They are equations where each term is either a constant or the product of a constant and a single variable, and they're characterized by their graph being a straight line.

• The general form of a linear equation in two variables is \( ax + by = c \).
• Linear equations can have one solution, no solutions, or infinitely many solutions, depending on the system setup.
• Systems of linear equations often appear in the form \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \).

Understanding the properties of linear equations helps in choosing the right method to solve a system, such as substitution or elimination, depending on the scenario and given data.

Substitution Method

The substitution method is another powerful strategy to solve systems of linear equations. Unlike elimination, which combines equations to eliminate variables, substitution takes a different approach by isolating one variable and substituting it back into the other equations.

• Select an equation and solve for one of the variables in terms of the others.
• Substitute this expression into another equation to eliminate one variable. This results in a single-variable equation.
• Solve this simplified equation for the remaining variable.
• Substitute back to find the other variable(s).

In our exercise, after using the elimination method to reduce the complexity, we used substitution to quickly find values for \( x \) and \( y \), leading us to the complete solution.

Solution Verification

Even after finding a solution to a system of equations, it's crucial to verify these values ensure they're accurate and indeed satisfy all original equations. Verification ensures that no mistakes were made during elimination or substitution.

• Substitute the found values for each variable back into the original equations.
• Verify that each equation balances, meaning both sides of each equation are equal with the substituted values.
• If any equation doesn't balance, re-check the work for potential algebraic mistakes.

In our problem, substituting \( x = \frac{1}{3} \), \( y = 0 \), and \( z = -2 \) back into the original equations confirms the solution, as every equation holds true, reinforcing the solution's validity.